Answer:
$208
Explanation:
Using the FIFO Inventory method, inventory items are assumed to be sold in the order in which they were purchased from the earliest to the latest.
The order of purchase of the inventory items are.
Jun. 1, DVD Player 1012, $113
Nov. 1, DVD Player 1045, $95
Nov. 31, DVD Player 1056, $88
Therefore, if two of the three items are sold, the cost of goods sold is the cost of the first two items purchased
= 113 + 95 = $208.
<span>An ATM card is used on January 24 to withdraw cash. The balance would be the remaining amount in the account once the withdrawal has been made. </span>
Answer:
- <u><em>It is best for Jerilyn to use the $10 coupon when the value of the purchase is equal or lower than $66.67, and it is best to use the $10 coupon when the value of the purchase is greater than $66.67</em></u>
Explanation:
Assume the value of the purchase is P.
Then <em>15%</em> of P is 0.15P.
To obtain the maximum benefit from the <em>15% coupon</em>, <em>Jerilyn</em> should use it when the discount from it is greater than the discount from the $10 coupon. This is:
Divide both sides by 0.15:
If the value of the purchase is equal to $66.67 the total discount with any cuopon are equal; if it is lower than $66.67, the discount of the $10 coupon is greater.
Thus, you conclude that for a $66.67 purchase she should use the $10 cuopon and for a purchase greater than $66.67 she should used the 15% cuopon.
Answer:
A) No, total values and imports and exports should be included in the calculation of the GNP.
Explanation:
The gross national product (GNP) must include the value of all imports and exports including intermediary goods.
Intermediary goods are goods used in the production of final goods, e.g. wood used to build a house. Intermediary goods can sometimes be final goods depending what use will be given to them, e.g. a tire is an intermediary good in the production of a car but it is also a final good when you buy a new tire to replace an old tire.
Answer:
0.3516 g/mL
Explanation:
Solubility = Mass of dry solute/ Volume of aqueous solution
= 87.9g / 250 ml
= 0.3516 g/mL